# Analysis Help - 1

• Dec 1st 2008, 04:46 PM
pila0688
Analysis Help - 1
IV. 1. Explain why it is that if a series converges (pointwise) on the closed interval [a, b] and converges uniformly on the open interval (a, b), then the series must converge uniformly on the closed interval [a, b]. (Note that to answer this question you must provide a proof that the series converges uniformly at the endpoints of the interval.)

IV. 2. Give an example of a function series that converges (pointwise) at every point of (a, b), each summand is continuous at every point in [a, b], but the series does not converge at every point of [a, b]. Formally prove that your function series satisfies all these conditions.
• Dec 2nd 2008, 07:28 AM
CaptainBlack
Quote:

Originally Posted by pila0688
IV. 1. Explain why it is that if a series converges (pointwise) on the closed interval [a, b] and converges uniformly on the open interval (a, b), then the series must converge uniformly on the closed interval [a, b]. (Note that to answer this question you must provide a proof that the series converges uniformly at the endpoints of the interval.)

For the series $\{s_n(x)\}$ to converge uniformly to $s(x)$ on $(a,b)$ for all $\varepsilon>0$ there exists an $N_{\varepsilon}$ such that for all $n>N_{\varepsilon}$:

$|s_n(x)-s(x)|<\varepsilon$

and $N_{\varepsilon}$ is independednt of $x$.

As the series converges at $x=s$ and $x=b$, for all $\varepsilon>0$ there exist $N_{1,\varepsilon}$ and $N_{2,\varepsilon}$ such that for all $n>N_{1,\varepsilon}$:

$|s_n(a)-s(a)|<\varepsilon$

and for all $n>N_{2,\varepsilon}:$

$|s_n(b)-s(b)|<\varepsilon.$

Now for $\varepsilon>0$ let $M_{\varepsilon}=\max(N_{\varepsilon}$, $N_{1,\varepsilon},N_{2,\varepsilon}).$

Now for all $x$ in $[a,b]$ we have for all $n>M_{\varepsilon}:$

$|s_n(x)-s(x)|<\varepsilon$

which proves that $\{ s_n(x) \}$ is uniformly convergent on $[a,b]$

CB